48 REV. T. P. KIRKMAN ON THE REPRESENTATION 



JO, and gained one face, 3 edges, and 3 summits, in K and its 

 sides and angles. Thus the number 3 of new edges of P", is 

 equal to that of the added face and summits together. 



U Ivo, p carries away, with / complete edges, /+1 summits 

 of P", provided that the section passes through no summit of 

 P"; but if it passes through i of them, p carries away / entire 

 edges and / — i+l summits of the prismoid solid. In the for- 

 mer case, P" after section has lost / edges and /-j-l summits, 

 and gained one face, k edges, and k summits: in the other 

 case, it has lost I edges and I — i-f 1 summits, while it has 

 gained k new summits on its k divided edges, and ^+2 new 

 edges in its new face of A+2 angles. 



The edges added to P" are k-\-i — /, and the faces and 

 summits added are ^+1 — (/— ^+l)=A+^ — L Of course, i is 

 either 1 or 2, or else 0. 



Thus the increase in the number of edges in P" after the 

 first section, is equal to the increase of faces and summits 

 together; and in the same way it is proved that a like equality 

 holds after the second or the w^* section. 



Now the theorem is true of P'', before it is cut ; for it has 

 3e edges, e+2 faces, and 2e summits. It is consequently 

 true after any section, and thus true of P', the result of the 

 final one, and therefore of P, which differs from P' only by 

 the subtraction of/ edges and / summits. Thus Theo. I. is 

 proved of any polyedron P, and that geometrically, in a man- 

 ner which may probably be new. 



2. The question — how many w-edrons are there? — has 

 been asked, but it is not likely soon to receive a definite 

 answer. It is far from being a simple question, even when 

 reduced to the narrower compass — how many w-edrons are 

 there whose summits are all triedral? Let us attempt to 

 consider this, or at least the first step of this, problem. 



If we take such an w-edron P, having n faces, s triedral 

 summits, and e edges, and count the edges that meet in all 



